# Thread: Limit_Problem Ex_RP

1. ## Limit_Problem Ex_RP

$lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n$

2. Note that:

$\left(\frac{n+1}{2n}\right) \left(\frac{2n+1}{n+1}\right)$

$=\frac{2n+1}{2n}=1+\frac{1}{2n}$

$\lim_{n\to \infty}\left(\frac{2n+1}{2n}\right)^{n}=\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$

Notice the similarity to the famous 'e' limit?.

$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}=e$

So, what is the limit of $\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$

3. Hint :
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$

4. Originally Posted by Raoh
Hint :
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$

$e^{\frac{1}{2}}$

but, what about :

$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n$

5. Originally Posted by Raoh
Hint :
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=?$

$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

6. Originally Posted by Raoh
$\lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

therefore your limit must be $e^{\frac{1}{2}}$.

7. O.k but what about
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$

8. Originally Posted by gilyos
O.k but what about
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$
i believe
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$

9. $
\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n = 0
$

Why?

Intuition:

$
(\frac{1}{2} + \frac{1}{2n})
$

is always less than 1, and then raise it to any power greater than one, and it will get smaller.

10. Originally Posted by gilyos
O.k but what about
$\lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
Given the discussion in this thread, a bit of thought might have suggested writing $\left( \frac{1}{2} + \frac{1}{2n}\right)^n = \left( \frac{1}{2}\right)^n \left( 1 + \frac{1}{n}\right)^n$ from which the limit is obvious.